Question

Show that a₁, a₂ … , a_n , … form an AP where a_n is defined as below

(i) a_n = 3 + 4n

(ii) a_n = 9 − 5n

Also find the sum of the first 15 terms in each case.

Answer 1

(i) a_n = 3 + 4n

a₁ = 3 + 4(1) = 7

a₂ = 3 + 4(2) = 3 + 8 = 11

a₃ = 3 + 4(3) = 3 + 12 = 15

a₄ = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂ − a₁ = 11 − 7 = 4

a₃ − a₂ = 15 − 11 = 4

a₄ − a₃ = 19 − 15 = 4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

= 15 × 35

= 525

(ii) a_n = 9 − 5n

a₁ = 9 − 5 × 1 = 9 − 5 = 4

a₂ = 9 − 5 × 2 = 9 − 10 = −1

a₃ = 9 − 5 × 3 = 9 − 15 = −6

a₄ = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a₂ − a₁ = − 1 − 4 = −5

a₃ − a₂ = − 6 − (−1) = −5

a₄ − a₃ = − 11 − (−6) = −5

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

= −465